Answers to Modules B6-B7From Chapter 3

3.1

X1 = Number of 20-inch girls bicycles produced this weekX2 = Number of 20-inch boys bicycles produced this weekX3 = Number of 26-inch girls bicycles produced this weekX4 = Number of 26-inch boys bicycles produced this week

MAX 27X1 + 32X2 + 38X3 + 51X4

S.T. X1 + X3 200 (Min girls models)

X2 + X4 200 (Min boys models)12X1 + 12X2 + 9X3 + 9X4 4800 (Production minutes) 6X1 + 9X2 + 12X3 + 18X4 4800 (Assembly minutes) 2X1 + 2X2 500 (20-inch tires)

2X3 + 2X4 800 (26-inch tires)All X's 0

150 20-inch girls, 100 20-inch boys, 100 26-inch girls, 100 26-inch boys; profit = $16,150

3.2a.

X1 = Number of stoves produced weeklyX2 = Number of washers produced weeklyX3 = Number of electric dryers produced weeklyX4 = Number of gas dryers produced weeklyX5 = Number of refrigerators produced weekly

MAX 110X1 + 90X2 + 75X3 + 80X4 + 130X5

S.T. 5.5X1 + 5.2X2 + 5.0X3 + 5.1X4 + 7.5X5 4800 (Molding/pressing) 4.5X1 1200 (Stove assembly) 4.5X2 + 4.0X3 + 3.0X4 2400 (Washer/dryer assembly)

9.0X5 1200 (Refrigerator assembly) 4.0X1 + 3.0X2 + 2.5X3 + 2.0X4 + 4.0X5 3000 (Packaging)

All X's 0

266.6667 stoves, 448.7179 Washers, 133.33333 refrigerators; Profit = $87,051.28Fractional values are work in progress from one week to the next.

b. Add the following constraints:X2 - X3 - X4 = 0 (Washers = Dryers) X3 - X4 100 (E. Dryers G. Dryers + 100) -X3 + X4 100 (G. Dryers E. Dryers + 100)

266.6667 stoves, 227.1545 washers, 63.57724 electric dryers, 163.5772 gas dryers, 133.3333 refrigerators; profit = $84,965.04

3.3

X1 = the number of standard Z345’s produced weeklyX2 = the number of industrial Z345’s produced weeklyX3 = the number of standard W250’s produced weeklyX4 = the number of industrial W250’s produced weeklyX5 = the total number of products produced weekly

MAX 400X1 + 560X2 + 500X3 + 700X4

S.T. 25X1 + 46X2 + 16X3 + 34X4 2500 (zinc) 50X1 + 30X2 + 28X3 + 12X4 2800 (iron) X1 + X2 20 (Min Z345’s)

X1 + X2 + X3 + X4 - X5 = 0 (X5 definition) X2 + X4 - .50X5 0 (Industrial min.)

X1 + X2 - .75X5 0 (Max Z345’s) X3 + X4 - .75X5 0 (Max W250’s)

X1, X2, X3, X4, X5 0

a. Produce 22.93578 standard Z345’s, 22.93578 standard W250’s, 45.87156 industrial W250’s (fractional production quantities are work in progress carried over from one week to the next). Weekly profit = $52,752.29

b. 75% (Slack is 0 on that constraint.) If this restriction is loosened or eliminated, the weekly profit will increase.

c. The shadow price of zinc is $21.10091743, which is valid for an additional 492.1568627 pounds.

i) 100 pounds is worth $2110.09 > $1500; yes purchase 100 additional pounds.ii) $2110.09 < $2600, no, 100 additional pounds should not be purchased.iii) Cannot tell without resolving since 800 additional pounds is outside the range of

feasibility for the shadow price. Re-solving (not shown) with 3300 pounds of zinc gives a profit of $68,055.87. Since this a $15,303.87 increase, then, yes, the 800 additional pounds should be purchased.

3.4

X1 = amount invested in EAL stockX2 = amount invested in BRU stockX3 = amount invested in TAT stockX4 = amount invested in long term bondsX5 = amount invested in short term bondsX6 = amount invested in the tax deferred annuityX7 = the total amount invested in stocks only

MAX .15X1 + .12 X2 + .09X3 + .11X4 + .085X5 + .06X6

S.T. X1 + X2 + X3 + X4 + X5 + X6 = 50,000 (Total)

X6 10,000 (TDA) X1 + X2 + X3 - X7 = 0 (Stocks)

X3 -.25X7 0 (Min TAT) X4 + X5 - X7 0(Bond stock) X3 + X5 + X6 12,500 (Low %)

All X's 0

a. Invest in (See following screen):EAL $ 7,500TAT $ 2,500Long Term Bonds $30,000Tax Deferred Annuity $10,000

Total return: $5,250

b. Rate of return for this investment = ($5,250/$50,000) = 10.5% The rate of return for additional funds = shadow price for the total investment constraint (the first constraint above) = 11% which is valid to + (1E+30).

c. The return on EAL cannot fall below 12% from 15%; the return on BRU cannot increase above 15% from 12%; the rate on long term bonds cannot increase above 13.5% from 11%.

d. Shadow price for: Total investment = .11 -- each additional dollar invested will earn 11% Minimum invested in taxed deferred annuity = -0.15 -- $0.15 lost for each extra dollar

required to be invested in tax deferred annuities. Stock Definition -- a meaningless shadow price; the right hand side will not change. Minimum invested in TAT = -0.16 -- a $0.16 decrease in return for each extra dollar

required to be in the low risk stock above 25%. Bonds >= Stocks = 0 -- no change in return for requiring bond investment to exceed

stock investment by at least $1. Maximum invested in low yield investments = 0.1 -- $0.10 additional return for each

extra dollar allowed to be invested in investments with returns less than 10%.

3.5Unit Profit

X1 = Number of full comforters produced daily 19-3(.50)-55(.20) = 6.50X2 = Number of queen comforters produced daily 26-4(.50)-75(.20) = 9.00X3 = Number of king comforters produced daily 32-6(.50)-95(.20) = 10.00

MAX 6.50X1 + 9.00X2 + 10.00X3

S.T. 3X1 + 4X2 + 6X3 2,700 (Stuffing) 55X1 + 75X2 + 95X3 48,000 (Fabric) 3X1 + 5X2 + 6X3 3,000 (Cutting minutes) 5X1 + 6X2 + 8X3 12,000 (Sewing minutes)

All X's 120

a. 240 full, 312 queen, 120 king comforters; daily profit = $5,5682688 pounds of stuffing, 48,000 sq. ft. of fabric, 3000 cutting minutes, 4032 sewing minutes used.

b. Allowable decrease = .13636 -- Minimum selling price = $26-.14 = $25.86c. (i) $0 2688 -

(ii) $0.11 46,800 - 48,200(iii) $0.15 2880 - 3080(iv) $0 4032 -

d. (See worksheet No Minimum) --300 full, 420 queen, 0 king; profit = $5,730

3.7X1 = Number of Student models produced each weekX2 = Number of Plus models produced each weekX3 = Number of Net models produced each weekX4 = Number of Pro models produced each week

MAX 70X1 + 80X2 + 130X3 + 150X4

S.T. X3 100 (Contract)

.4X1 + .5X2 + .6X3 + .8X4 750 (Production Hours) X1 + + X3 700 (Celeron) X2 + X4 550 (Pentium) X1 + X2 + X3 + 800 (20gb Hard Drives)

X4 950 (30gb Hard Drives) X1 + X2 + 2X3 + X4 1600 (Floppy Drives) X1 + X2 + X4 1000 (Zip Drives) X1 + X3 + X4 1600 (CD R/W's) X2 + X3 + X4 900 (DVD's) X1 + X2 850 (15-in. monitors)

X3 + X4 800 (17-in. monitors) X2 + X3 1250 (Mini-tower cases)

X1 + X4 750 (Tower cases)All X's 0

Given the output shown on the next page:a. 325 Student models, 100 Plus models, 375 Net models, 425 Pro models; weekly

profit = $143,250

b. Since Allowable Decrease for the Plus model is 55, the optimal solution could change if its profit coefficient is reduced below $80 - $55 = $25. But this does not necessarily guarantee the production will now be 0, only that it could change. But if $24.99 is substituted for $80 and Solver is called again (not shown--see worksheet Plus Profit $24.99), there is no change to the optimal solution. Now the Allowable Decrease for the Plus model is $24.99, so the optimal solution could change if its profit coefficient is reduced below $24.99 - $24.99 = $0. Set the profit for the Plus model to -$0.01 and call Solver again. Now the optimal production quantity for the Plus model is 0. So, its profit can fall to $0, before it becomes unprofitable to produce the Plus model.

c. Yes, the shadow price for 17-inch monitors is $25.

d. No, production hours is not a binding constraint; DO NOT HIRE THE WORKER.

3.8 See file Ch3.8.xls

X1 = the number of Delta assemblies produced dailyX2 = the number of Omega assemblies produced dailyX3 = the number of Theta assemblies produced daily

MAX 800X1 + 900X2 + 600X3

S.T. X1 + X2 + X3 7 (X70686 chips) 2X1 + X2 + X3 8 (Production hours) 80X1 + 160X2 + 80X3 480 (Quality minutes)

All X's 0

From the output on the next page:

a. Produce 2 Delta’s and 4 Theta’s; daily profit = $4,000. No Omega’s are produced because the use too much quality control time compared to its slightly higher unit profit.

b. Reduced cost = -100; thus minimum profit for production = $1,000. Since costs are $900, this implies a minimum price of $1,900.

c. 6; there is a slack of 1.

d. (i) There is slack on X70686 chips, thus additional profit is $0. (ii) Production hours are sunk costs. Since the shadow price for production hours is $200 and 3 additional hours is within the range of feasibility, these 3 hours gross 3($200) = $600 additional profit. Thus the net additional profit = $600 - $525 = $75. (iii) Quality control hours are also sunk costs. Since 1 additional hour = 60 additional minutes is also within its range of feasibility, the $5 shadow price is valid for all 60 minutes grossing 60($5) = $300 additional profit. The net additional profit is $300 - $200 = $100.

OPTION (iii) is of the most value.

3.10

X1 = the number of ounces of Multigrain Cheerios in the mixtureX2 = the number of ounces of Grape Nuts in the mixtureX3 = the number of ounces of Product 19 in the mixtureX4 = the number of ounces of Frosted Bran in the mixtureX5 = the total number of ounces in the mixture

MIN 12X1 + 9X2 + 9X3 + 15X4

S.T. 30X1 + 30X2 + 20X3 + 20X4 50 (Vitamin A)25X1 + 2X2 + 100X3 + 25X4 50 (Vitamin C)25X1 + 25X2 + 25X3 + 25X4 50 (Vitamin D)25X1 + 25X2 + 100X3 + 25X4 50 (Vitamin B6)45X1 + 45X2 + 100X3 + 25X4 50 (Iron) X1 + X2 + X3 + X4 - X5 = 0 (Total) X1 - .1X5 0 ( 10% M/G Cheerios)

X2 - .1X5 0 ( 10% Grape Nuts) X3 - .1X5 0 ( 10% Product 19)

X4 - .1X5 0 ( 10% Frosted Bran) All X's 0

a. From output below: Total sugar intake = 19.8 grams with this combination:

.2 oz. Multigrain Cheerios1.2244898 oz.Grape Nuts

.3755102 oz. Product 19 .2 oz.Frosted Bran

Total 2.0 oz.

The Allowable Increase for the sugar coefficient for Grape Nuts is 0, and the Allowable Decrease for the sugar coefficient for Product 19 is 0. This indicates alternate optimal solutions. By setting cell H6 to 19.8, then changing the target cell to MIN D4 (on worksheet Alternate (not shown)), another optimal solution is:

.2 oz. Multigrain Cheerios

.8 oz. Grape Nuts

.8 oz. Product 19

.2 oz. Frosted BranTotal 2.0 oz.

b. Total = 2 oz. of cereal and 2 (1/2) = 1 cup of skim milk

c. Extra % above 50% required of vitamin D adds .396 grams of sugar Extra ounce above 10% required of Multigrain Cheerios adds 3 grams of sugar Extra ounce above 10% required of Frosted Bran adds 6 grams of sugar

4.6d. Product 19 has less sugar and gives percentages that are at least as large as those for Frosted Bran for every vitamin and iron requirement. Re-solving gives the following alternate optimal solutions with 18 grams of sugar.

Mixture 1 (Shown) Mixture 2 (On Worksheet Alternate No 10%)1.53 oz. Grape Nuts 1.0 oz. Grape Nuts .47 oz. Product 19 1.0 oz. Product 19

Total 2.00 oz. Total 2.0 oz.

3.11

X1 = Number of refrigerator/ovens producedX2 = Number of French fry makers producedX3 = Number of French toast makers produced

MIN 140X1 + 50X2 + 36X3 S.T.

100X1 + 35X2 + 27X3 2,000,000 (Min Profit) X1 5,000 (Min Refrig/oven)

X2 4,000 (Min French fry maker) X3 2,300 (Min French toast maker)

X1 15,000 (Max Refrig/oven) X2 15,000 (Max French fry maker)

X3 15,000 (Max French toast maker)

Make 14,550 refrigerator ovens, 4000 French fry makers, 15,000 French toast makersTotal Variable Cost = $2,777,000.

3.12a.

X1 = Number of plates made per dayX2 = Number of mugs made per dayX3 = Number of steins made per dayX4 = Total daily production

MAX 2.50X1 + 3.25X2 + 3.90X3

S.T. 2X1 + 3X2 + 6X3 1920 ((4)(8)(60) Molding min.) 8X1 + 12X2 + 14X3 3840 ((8)(8)(60) Finishing min.) X2 150 (Minimum mugs) -2X1 - 2X2 + X3 0 (Steins 2(Plates + Mugs) X1 + X2 + X3 - X4 = 0 (Total Definition) X1 - .3X4 0 (Plates 30% Total

Produced) All X's 0

101.8033 plates, 150 mugs, 87.54098 steins; total daily profit = $1083.42

b. Combine the first two constraints into one:

10X1 + 15X2 + 20X3 5760

128 plates, 298.6667 mugs, 0 steins; total daily profit = $1290.67

This is an increase of ($1290.67 - $1083.42) = $207.25

3.15

X1 = the number of acres of wheat plantedX2 = the number of acres of corn plantedX3 = the number of acres of oats plantedX4 = the number of acres of soybeans planted

Profit coefficients are 210($3.20) - $50 = $622, 300($2.55) - $75 = $690, 180($1.45) - $30 = $231, and 240($3.10) - $60 = $684 respectively.

MAX 622X1 + 690X2 + 231X3 + 684X4

S.T. 4X1 + 5X2 + 3X3 + 10X4 1,800 (Labor hours) 50X1 + 75X2 + 30X3 + 60X4 25,000 (Expenses) 2X1 + 6X2 + X3 + 4X4 1,200 (Water)210X1 30,000 (Min. Wheat)

300X2 30,000 (Min. Corn) 180X3 25,000 (Max Oats)

X1 + X2 + X3 + X4 300 (Total acres)All X's 0

a. (See Worksheet on next page) Plant 142.8571 acres of wheat, 142.8571 acres of corn and 14.28571 acres of soybeans; profit = $197,200.

b. The net profit must rise to $675; adding the $30/acre in expenses = $705. If selling price remains $1.45 per bushel, then yield must increase to $705 per acre/$1.45 per bushel = 486.21 bushels per acre.If the yield remains 180 bushels per acre, then its price must rise to $705 per acre/180 bushels per acre = $3.92 per bushel.

c. (See work sheet NO CORN (not shown).) Yes, corn would still be planted; there is currently slack on the corn constraint. If corn were not grown, the problem must be re-solved. Plant 200 acres of wheat and 100 acres of soybeans for a profit of $192,800 -- a decrease of $4,400.

d. The range of feasibility for acres is only valid up to 318.57 acres (18.57 additional acres.) Thus the problem must be re-solved. See worksheet PARCEL.

Plant 181.82 acres of wheat, 101.82 acres of corn, and 56.36 acres of soybeans for a profit of $221,898.18 -- an increase of $24,698.18. Yes Bill should lease this property for $2,000.

3.18

X1 = amount invested in Bonanza GoldX2 = amount invested in Cascade TelephoneX3 = amount invested in the money market accountX4 = amount invested in two-year treasury bonds

MAX .15X1 + .09X2 + .07X3 + .08X4

S.T. X1 + X2 50,000 (Max stocks) X1 + X2 + X3 60,000 (Min potential 9%) X4 30,000 (Max treasury bonds)-.5X1 + .03X2 + .06X3 + .08X4 4,000 (Min return) X1 + X2 + X3 + X4 = 100,000 (Total)

All X's 0

Invest $2,075.47 in Bonanza Gold, $47,924.53 is Cascade Telephone, $20,000 in the money market account, $30,000 in two year treasury bonds. The expected return is $8,424.53 (8.42453%).

3.19

Xij = number of gallons of crude i blended into grade ji = p (Pacific), g (Gulf), m (Middle East) j = r (Regular), p (Premium)

Xj = total amount of grade j produced

Example of profit coefficient: Selling price of regular = $0.52, purchase cost of Pacific crude = ($14.28)/42 = $0.34; thus profit on a gallon of Pacific crude = $0.52 - $0.34 = $0.18

MAX .18Xpr +.16Xgr +.05Xmr +.26Xpp +.24Xgp +.13Xmp

S.T. Xpr + Xpp 126,000 (Pacific) Xgr + Xgp 84,000 (Gulf)

Xmr + Xmp 336,000 (M. East) Xpr + Xgr + Xmr - Xr = 0 (Regular)

Xpp + Xgp + Xmp – Xp = 0 (Premium) Xr 200,000 (Min Reg)

Xp 100,000 (Min Prem) Xr + Xp 400,000

(Capacity)85Xpr + 87Xgr + 95Xmr - 87Xr 0 (Reg Oct.)

85Xpp +87Xgp + 95Xmp - 91Xp 0 (Prem Oct.) All X's 0

From the sensitivity report (not shown), there are alternate optimal solutions giving a profit of $61,620. The one represented on the screen on the next page is:

Pacific - Regular 46,000Gulf - Regular 84,000Mid East - Regular 70,000Pacific - Premium 80,000Gulf - Premium 0Mid East - Premium 120,000

b. Shadow price (from Total Total (not shown)) = $0.13 per gallon; Thus 50,000 gallons is worth 50,000($0.13) = $6,500 > $5,000 -- Yes Caloco should secure the extra refining capacity.

c. The solution uses all 190,000 gallons of Mid East Oil @ $0.47/gallon = $89,300. 8000 barrels @ $16.80 per barrel = $134,400 -- the Middle East distributors receive

more.To see if it would be profitable for Caloco, the problem must be re-solved.

$16.80 per barrel = $0.40 per gallon. The new profit coefficients for Xmr and Xmp would be $0.12 and $0.20 respectively. Change the third constraint to Xmr + Xmp = 336,000 Re-solve (See worksheet Mid East (not shown).)The new solution gives Caloco a profit of $67,840 -- thus it would be profitable for Caloco to accept this offer. There is now no Gulf crude purchased and only slightly more

than 1500 barrels of Pacific crude purchased. These distributors would have to cut their prices to stay competitive.

Minimum

AvailabilityCapacit

Octane

3.23a. This could violate the “no interaction” assumption of linear programming.

b. X1 = $ spent on television advertisingX2 = $ spent on radio advertisingX3 = $ spent on newspaper advertisingX4 = $ spent on its circulars

MAX 28X1 + 18X2 + 20X3 + 15X4

S.T. X1 + X2 + X3 + X4 700,000 (Budget) X1 + X2 350,000 (TV/Radio)10X1 + 7X2 + 8X3 + 4X4 2,500,000 (Yuppies) 5X1 + 2X2 + 3X3 + X4 1,200,000 (College) 5X1 + 8X2 + 6X3 + 9X4 1,800,000 (Audiophiles) X1 300,000 (Max TV)

X2 300,000 (Max Radio) X3 300,000 (Max Newspapers)

X4 300,000 (Max Circulars)All X's 0

$300,000 TV, $100,000 radio, $300,000 newspapers Total Exposure = 16,200,000

c. It would not affect the optimal solution; it is a non-binding constraint (there is surplus).

3.26

X1 = the number of 2-oz. Go bars produced dailyX2 = the number of 2-oz. Power bars produced dailyX3 = the number of 2-oz. Energy bars produced dailyX4 = the number of 8-oz. Energy bars produced dailyX5 = the total number of 2-oz. bars produced daily

Derivation of profit coefficients The costs per 2oz. are: Protein concentrate $0.40; Sugar substitute $0.175; Carob $0.325.Thus the unit profits are:

2 oz. Go bars: .68 - .03 - .2(.40) - .6(.175) - .2(.325) = $0.402 oz. Power bars: .84 - .03 - .5(.40) - .3(.175) - .2(.325) = $0.49252 oz. Energy bars: .76 - .03 - .3(.40) - .4(.175) - .3(.325) = $0.44258 oz. Energy bars: 3.00 - .05 - .3(1.60) - .4(.70) - .3(1.30) = $1.80

MAX .40X1 + .4925X2 + .4425X3 +1.80X4

S.T. .4X1 + X2 + .6X3 + 2.4X4 9,600 (Oz. protein)1.2X1 + .6X2 + .8X3 + 3.2X4 16,000 (Oz. sugar sub.) .4X1 + .4X2 + .6X3 + 2.4X4 12,800 (Oz. carob) X1 + X2 + X3 - X5 = 0 (Total 2-oz.)

X5 25,000 (Max 2-oz.) X4 2,000 (Max 8-oz.)

X1 2,500 (Min 2-oz. Go) X2 2,500 (Min 2-oz. Power)

X3 2,500 (Min 2-oz. Energy) X1 -.5X5 0 (Max 2-oz. Go)

X2 -.5X5 0 (Max 2-oz. Power) X3 -.5X5 0 (Max 2-oz. Energy) 2X3 + 4X4 - X5 0 (Max Energy*)

All X's 0

*The Maximum Energy Bar constraint is formulated as follows: Total Weight Energy Bars: 2X3 + 8X4 Total Weight All Bars: 8X4 + 2X5

Thus, 2X3 + 8X4 .5(8X4 + 2X5) or 2X3 + 4X4 - X5 0.

Daily production, giving daily profit of $7,273.375 (see output on next page): 2-oz. Go: 7,550

2-oz. Power: 2,5002-oz. Energy: 5,0508-oz. Energy 437.5

3.28Xj = the number of workers that have shift j

MIN 15X1 + 25X2 + 52X3 + 22X4 + 54X5 + 24X6 + 55X7 + 23X8 + 16X9

S.T. X1 + X2 + X3 8 X2 + X3 10

X3 + X4 + X5 22 X3 + X4 + X5 15

X5 + X6 + X7 10 X5 + X6 + X7 20

X7 + X8 16 X7 + X8 + X9 8

X3 2 X7 2

.6X3 - .4X4 + .6X5 0 .6X5 - .4X6 + .6X7 0All X’s 0, and integer

Shift 2--7, Shift 3--3, Shift 4--13, Shift 5--6, Shift 6--12, Shift 7--2, Shift 8--14 Total Cost = $1,661.

3.31 NOTE: Change Tolerance in Options Dialogue box to .5%.

X1 = the number of Tropic homes builtX2 = the number of Sea Breeze homes builtX3 = the number of Orleans homes builtX4 = the number of Grand Key homes built

MAX 40,000X1 + 50,000X2 + 60,000X3 + 80,000X4

S.T. .20X1 + .27X2 + .22X3 + .35X4 20 (Acres) X1 + X2 40 (One story) X2 + X3 + X4 50 (3+ BR) X1 10 (Min Trop.)

X2 10 (Min Sea Br.)X3 10 (Min Orleans)

X4 10 (Min Gr. Key)All X's 0 and integer

a. Build 29 Tropic, 11 Sea Breeze, 35 Orleans, 10 Grand Key models; profit = $4,610,000

b.

Change to .5

30 Tropic, 10 Sea Breeze, 35 Orleans, 10 Grand Key models; profit = $4,600,000. This solution satisfies all the constraints but is $10,000 less than the optimal solution.

c. Let Yi = 1 if the constraint holds and Yi = 0 if it does not

Add the following constraints:X1 - 12Y1 0X2 - 12Y2 0X3 - 12Y3 0X4 - 12Y4 0Y1 + Y2 + Y3 + Y4 3

NOTE: Excel may incorrectly print that the problem is infeasible. But the solution below is feasible and optimal.

Build 30 Tropic, 10 Sea Breeze, 32 Orleans, 12 Grand Key models; profit = $4,580,000. Note: There are alternate optimal solutions.

3.32 In Options dialogue box, Change tolerance to .5% and check Use Automatic Scaling

X1 = the number of Nissan vans Logitech should purchaseX2 = the number of Toyota vans Logitech should purchaseX3 = the number of Plymouth vans Logitech should purchaseX4 = the number of Ford stretch vans Logitech should purchaseX5 = the number of Mitsubishi minibuses Logitech should purchaseX6 = the number of General Motors minibuses Logitech should purchaseX7 = the total number of vehicles Logitech should purchaase

MAX 7X1 + 8X2 + 9X3 + 11X4 + 20X5 + 24X6

S.T. 26000X1 + 30000X2 + 24000X3 + 32000X4 + 50000X5 + 60000X6 250,000 5000X1 + 3500X2 + 6000X3 + 8000X4 + 7000X5 + 11000X6 50,000

X1 + X2 + X3 + X4 + X5 + X6 - X7 = 0 X7 8

X5 + X6 1 X1 + X2 + X3 + X4 3

X3 + X4 + X6 -.5X7 0 All X’s 0, and integer

Change to .5

CHECK

a. Maximum Capacity = 97 using 2 Plymouth vans, 1 Ford van, 1 Mitsubishi minibus, 2 General Motors minibuses.

b. See worksheets b-253900, b-254000, b-249900, b-259900, b-260000 (not shown)

Capacity Nissan Toyota Plymouth Ford Mitsubishi General Motors(i) 97 2 1 1 1(ii) 98 3 1 3(iii) 96 4 3(iv) 100 4 2 1 (v) 100 4 2 1

Sensitivity of the right hand side gives non-smooth jumps to new optimal solutions.

c. The problem is infeasible. The minimum van and mini-bus constraints require a minimum budget of $122,000.

3.34

a. Xj = the number of software application j developed (j = 1, 2, 3, 4, 5, 6)

MAX 2X1 + 3.6X2 + 4X3 + 3X4 + 4.4X5 + 6.2X6 (in $millions)S.T. 6X1 + 18 X2 + 20X3 + 16X4 + 28X5 + 34X6 60 (Programmers) .4X1 + 1.1X2 + .94X3 + .76X4 + 1.26X5 + 1.8X6 3.5 (Budget in $millions)

All Xj’s binary

Korvex should develop applications 1, 2, 3, and 4; net present worth = $12,600,000.

b. Add the following constraints to the formulation in part a.

X4 - X5 = 0 (Proj 4 = Proj 5)- X1 + X2 0 (Proj 2 only if Proj 1)

X3 + X6 1 (Not both Proj 3 and Proj 6) X1 + X2 + X3 + X4 + X5 + X6 3 (Max 3 applications)

See worksheet Part b (not shown).Korvex should develop applications 1, 2, and 6; net present worth = $11,800,000.

3.35

X1 = the number of CPA’s hiredX2 = the number of experienced accountants hiredX3 = the number of junior accountants hired

The total number of accounts that can be serviced is 6X1 + 6X2 + 4X3. This must be greater than or equal to 100 plus the number of corporate accounts serviced: 6X1 + 6X2 + 4X3 100 + (3X1 + X2); and the number of corporate accounts serviced must be at least 25: 3X1 + X2 25. Also the number of CPA’s and experienced accountants (X1 + X2) must be at least two-thirds of all employees hired (X1 + X2 + X3) or X1 + X2 2/3(X1 + X2 + X3). Rearranging terms gives the following:

MIN 1200X1 + 900X2 + 600X3

S.T. 3X1 + 5X2 + 4X3 100 (Service 100 personal accounts) 3X1 + X2 25 (Service 25 corporate accounts) 1/3X1 + 1/3X2 - 2/3X3 0 ( 2/3 are CPA’s or experienced) All X's 0 and integer

Hire 2 CPA’s and 19 experienced accountants; total payroll = $19,500.

3.36

X1 = the number of TV exposures X2 = the number of radio exposuresX3 = the number of newspaper exposures

a. MAX 500,000X1 + 50,000X2 + 200,000X3

S.T. X1 250 (Max TV) X2 250 (Max radio)

X3 250 (Max newspapers) 4,000X1 + 500X2 + 1,000X3 500,000 (Budget)

All X's 0 and integer

Use 62 TV exposures, 4 radio exposures, and 250 newspaper exposures; total audience reached = 81,200,000.

b. Change the objective function to: MIN 4,000X1 + 500X2 + 1,000X3 and change the third constraint to: 500,000X1 + 50,000X2 + 200,000X3 30,000,000. See worksheet Century -- Min Cost (not shown). Use 150 newspaper exposures only; total cost $150,000.

c.

MAX 500,000X1 + 50,000X2 + 200,000X3

S.T. X1 - 250Y1 0 X2 - 250Y2 0 X3 - 250Y3 0

4,000X1 + 500X2 + 1,000X3 + 500,000Y1 + 50,000Y2 + 100,000Y3 1,000,000All X's 0 and integer All Y's binary

Use 38 TV exposures, 0 radio exposures, and 248 newspaper exposures; total audience reached = 68,600,000.